Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the m
1 answer:
You might be interested in
Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :
So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer:
The speed at which the ball rolled off the end of the table is 3.3 m/s
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.
The position vector of the ball can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity.
When the ball reaches the ground, its position will be:
r final = (3, -4)
Then:
3 = x0 + v0x · t
-4 = y0 + v0y · t + 1/2 · g · t²
Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:
3 m = v0x · t
-4 m = 1/2 · g · t²
We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:
-4 m = 1/2 · g · t²
-4 m = -1/2 · 9.8 m/s² · t²
8 m / 9.8 m/s² = t²
t = 0.9 s
Then:
3 m = v0x · 0.9s
3 m/ 0.9 s = v0x
v0x = 3.3 m/s
The speed at which the ball roll off the end of the table is 3.3 m/s
